1

Color Superconductivity: Color Superconductivity: CFL and 2SC phasesCFL and 2SC phases

•• IntroductionIntroduction

•• Hierarchies of effective Hierarchies of effective lagrangianslagrangians

•• Effective theory at the Fermi surface (HDET)Effective theory at the Fermi surface (HDET)

•• Symmetries of the superconductive phasesSymmetries of the superconductive phases

2

IntroductionIntroductionIdeas about CS back in 1975 (Collins & PerryIdeas about CS back in 1975 (Collins & Perry--1975, 1975,

BarroisBarrois--1977, Frautschi1977, Frautschi--1978).1978).

Only in 1998 (Alford, Only in 1998 (Alford, RajagopalRajagopal & & WilczekWilczek; Rapp, ; Rapp, Schafer, Schafer, SchuryakSchuryak & & VelkovskyVelkovsky) a real progress) a real progress..

The phase structure of QCD at highThe phase structure of QCD at high--density density depends on the number of flavors with massdepends on the number of flavors with mass m <m < µµ..

TwoTwo most interesting cases:most interesting cases: NNff = 2, 3= 2, 3..

Due to asymptotic freedom quarks are almost free Due to asymptotic freedom quarks are almost free at high density and we expect at high density and we expect difermiondifermioncondensation in the color channel condensation in the color channel 33**..

3

Consider the possible pairings at very high densityConsider the possible pairings at very high density

α βia jb0 ψ ψ 0 α, β α, β color; color; i, j i, j flavor;flavor; a,ba,b spinspin

AntisymmetryAntisymmetry in spinin spin ((a,ba,b) ) for better use of for better use of the Fermi surfacethe Fermi surface

AntisymmetryAntisymmetry in colorin color ((α, βα, β) ) for attractionfor attraction

AntisymmetryAntisymmetry in flavorin flavor ((i,ji,j) ) for for PauliPauli principleprinciple

4

pp−

ss− Only possible pairings Only possible pairings

LL and RRLL and RR

For For µ >> µ >> mmuu, , mmdd, m, mss

Favorite state forFavorite state for NNff = 3, = 3, CFLCFL (color(color--flavor flavor locking) (Alford, locking) (Alford, RajagopalRajagopal & & WilczekWilczek 1999)1999)

α β α β αβCiL jL iR jR ijC0 ψ ψ 0 = - 0 ψ ψ 0 ∆ε ε∝

Symmetry breaking patternSymmetry breaking pattern

c L R c+L+RSU(3) SU(3) SU(3) SU(3)⊗ ⊗ ⇒

5

Why CFL?Why CFL? α β αβCiL jL ijC0 ψ ψ 0 ∆ε ε∝

α β αβCiR jR ijC0 ψ ψ 0 ∆ε ε∝ −

6

What happens going down with What happens going down with µµ? If? If µ µ << m<< mss, , we get we get

3 colors and 2 flavors (2SC)3 colors and 2 flavors (2SC)α β αβ3iL jL ij0 ψ ψ 0 = ∆ε ε

c L R c L RSU(3) SU(2) SU(2) SU(2) SU(2) SU(2)⊗ ⊗ ⇒ ⊗ ⊗

However, if However, if µ µ is in the intermediate region is in the intermediate region we face a situation with fermions having we face a situation with fermions having different Fermi surfaces (see later). Then different Fermi surfaces (see later). Then other phases could be important (LOFF, etc.)other phases could be important (LOFF, etc.)

7

Difficulties with lattice calculationsDifficulties with lattice calculations

•• Define Define eucldeaneucldean variables:variables: 4 i i0 E E

4 i i0 E E

x ix , x x

, i

→ − →

γ → γ γ → − γ

•• DiracDirac operator with chemical potentialoperator with chemical potential4

E E ED( ) Dµ µµ = γ + µγ

•• At At µ = 0 µ = 0 †5 5D(0) = -D(0), γ D(0)γ = -D(0)

•• EigenvaluesEigenvalues of of D(0)D(0) pure pure immaginaryimmaginary

•• If eigenvector of D(0), If eigenvector of D(0), eigenvector with eigenvector with eigenvalueeigenvalue -- λλ

λ 5γ λ

8

det[D(0)] ( )( ) 0= λ −λ >∏

For For µ µ not zeronot zero the argument does not apply the argument does not apply and one cannot use the sampling method for and one cannot use the sampling method for evaluating the determinant. However for evaluating the determinant. However for isospinisospin chemical potential and two degenerate chemical potential and two degenerate flavors one can still prove the flavors one can still prove the positivitypositivity..

For finite baryon density no lattice For finite baryon density no lattice calculation available except for small calculation available except for small µ µ and and close to the critical line (Fodor and Katz)close to the critical line (Fodor and Katz)

9

10

Hierarchies of effective Hierarchies of effective lagrangianslagrangians

Integrating out Integrating out heavy degrees of heavy degrees of

freedom we have two freedom we have two scales. The gap scales. The gap ∆∆ and and

a cutoff, a cutoff, δδ above above which we integrate which we integrate

out. Therefore: out. Therefore: two different two different

effective theories, effective theories, LLHDETHDET and and LLGoldsGolds

11

•• LLHDETHDET is the effective theory of the is the effective theory of the fermions close to the Fermi surface. It fermions close to the Fermi surface. It corresponds to the corresponds to the PolchinskiPolchinski description. description. The condensation is taken into account by The condensation is taken into account by the introduction of a mean field the introduction of a mean field corresponding to a corresponding to a MajoranaMajorana mass. mass. The The d.o.fd.o.f. are quasi. are quasi--particles, holes and gauge particles, holes and gauge fields. fields. This holds for energies up to the This holds for energies up to the cutoffcutoff..

•• LLGoldsGolds describesdescribes the low energy modes (the low energy modes (E<< E<< ∆∆), as ), as Goldstone bosons, Goldstone bosons, ungappedungapped fermions fermions and holes and and holes and masslessmassless gauge fieldsgauge fields, , depending on the breaking scheme.depending on the breaking scheme.

12

Effective theory at the Effective theory at the Fermi surface (HDET)Fermi surface (HDET)

Starting point: Starting point: LLQCDQCD

a aQCD 0

a a a as

1L iD F F , a 1, ,84

D ig A T , D D , T2

µνµν

µµ µ µ µ

= ψ ψ − + µψγ ψ = ⋅⋅ ⋅

λ= ∂ + = γ =

at asymptotic at asymptotic µ >> Λµ >> ΛQCDQCD,,

0 0( p ) (p) 0 (p ) (p) p (p)+ µγ ψ = + µ ψ =⇒ α ⋅ ψ

0 2 2 0(p ) | p | p E | p |±⇒+ µ = = = −µ±

13

Introduce the projectors:Introduce the projectors:

F F

FF

p p p p

1 v E(p) | p | ˆP , v v p2 p p±

= =

± α ⋅ ∂ ∂= ≡ = = =

∂ ∂

Fp v= µ +and decomposing:and decomposing:

Fvµ

H

H ( 2 )+ +

− −

ψ = α ⋅ ψ

ψ = − µ + α ⋅ ψ

•• States States ψψ+ + close to the FS close to the FS

•• States States ψψ−− decouple for large decouple for large µµ

14

FieldField--theoretical version:theoretical version:4

ip x4

d p(x) e (p)(2 )

− ⋅ψ = ψπ∫

0

p v

v (0, v), v 1, ( , )

v , ( v)v

µ µ µ

µ µ

⊥ ⊥

= µ +

= = =

= + = − ⋅

v p

0⊥ =0, , v

ChoosingChoosing

4 4 d.ofd.of. .

Separation of light and heavy Separation of light and heavy d.o.fd.o.f..

light d.o.f .heavyd.o.f

| p | ,| p | ,| p | ,. ,

µ − δ ≤ ≤ µ + δ

≤

− δ ≤ ≤ +

µ − δ ≥ µ

δ

≤ −δ ≥ +δ+ δ

15

Separation of light and heavy Separation of light and heavy d.o.fd.o.f..

heavyheavy

heavyheavy

lightlightlightlight

δ

16

Momentum integration for the light fieldsMomentum integration for the light fields

4 2 2 2

04 4 2d p dv dd d d

(2 ) (2 ) 4 (2 )

+δ +∞

−δ −∞

µ µ⇒ Ω =

π π π π π∫ ∫ ∫ ∫ ∫ ∫The Fourier decomposition becomesThe Fourier decomposition becomes

vv11

vv22

i v xv

dv x)4

((x) e− µ ⋅ψ = ψπ∫2 2

i xvv 2

d e ( )(2 )

(x) − ⋅µ= ψ

π πψ ∫( )v ( ) (p)ψ ≡ ψ

For any fixed For any fixed v, v, 22--dim theorydim theory

17

In order to decouple the states corresponding In order to decouple the states corresponding to to E−

[ ]i v x

2 2i x

v v2

dv(x) e (x) (x)4

d(x) P (x) P e ( )(2 )

− µ ⋅+ −

− ⋅± ± ±

ψ = ψ + ψπ

µψ ≡ ψ = ψ

π π

∫

∫

MomentaMomenta from the Fermi spherefrom the Fermi sphere

substituting inside substituting inside LLQCD QCD and usingand using

2 24 † †

v v2dv dd x (x) (x) (x) (x)4 (2 )µ→∞

µψ ψ → ψ ψ

π π π∫ ∫

18

4 † †0

dvd x (iD ) iV D (2 iV D) ( iD h.c.)4 + + − − + ⊥ − ψ + µγ ψ → ψ ⋅ ψ + ψ µ + ⋅ ψ + ψ ψ + π∫ ∫

V (1, v), V (1, v)µ µ= = −

0( , (v )v),µ µ µ µ⊥γ = γ ⋅ γ γ = γ − γ

†

†

V

V

µ µ+ + + +

µ µ− − − −

µ µ+ − + ⊥ −

µ µ− + − ⊥ +

ψ γ ψ = ψ ψ

ψ γ ψ = ψ ψ

ψ γ ψ = ψ γ ψ

ψ γ ψ = ψ γ ψ

0

0

iV D i D 0

(2 iV D) i D 0+ ⊥ −

− ⊥ +

⋅ ψ + γ ψ =

µ + ⋅ ψ + γ ψ =EqsEqs. of motion:. of motion:

19

iV D 00

+

−

⋅ ψ =ψ =

At the leading At the leading order in order in µ:µ:

At the same order:At the same order:

†D

dvL iV D4 + += ψ ⋅ ψ

π∫Propagator:Propagator: 0

02 2

0 0

0 0 †

1 (p ) p V 1 ( T( ) )p (p ) | p | 2 V

V 1 (1 v) P ( T( ) )2 2

µ→∞

+

+ µ γ − ⋅ γ= → ψψ

+ µγ + µ − ⋅

= γ − α ⋅ = γ ψ ψ1

V ⋅

20

Integrating out the heavy Integrating out the heavy d.o.fd.o.f..

For the heavy For the heavy d.o.fd.o.f. we can formally . we can formally repeat the same steps leading to:repeat the same steps leading to:

4 † †0

dvd x (iD ) iV D (2 iV D) ( iD h.c.)4 + + − − + ⊥ − ψ + µγ ψ → ψ ⋅ ψ + ψ µ + ⋅ ψ + ψ ψ + π∫ ∫

Eliminating the Eliminating the EE-- fields one would get the fields one would get the nonnon--local local lagrangianlagrangian::

† †D

dv 1L iV D P D D4 2 iV D

1P g V V V V2

µν+ + + µ ν +

µν µν µ ν ν µ

= ψ ⋅ ψ − ψ ψ π µ + ⋅

= − +

∫

21

h+ + +ψ = ψ + ψDecomposingDecomposing one getsone gets

h hD D D D

†D

h † h † hD

h h† h h† hD

L L L LdvL iV D4

dv 1L iV D P D D h4 2 iV D

dv 1L iV D P D D4 2 iV D

+ +

µν+ + + µ ν +

µν+ + + µ ν +

= + +

= ψ ⋅ ψπ

= ψ ⋅ ψ − ψ ψ + ↔ π µ + ⋅

= ψ ⋅ ψ − ψ ψ π µ + ⋅

∫

∫

∫

When integrating out the heavy fieldsWhen integrating out the heavy fields

22

Contribute only if some gluons Contribute only if some gluons are hard, but suppressed by are hard, but suppressed by

asymptotic freedomasymptotic freedom

From From LLhhDD gives the bare gives the bare MeissnerMeissner massmass

23

HDET in the condensed phaseHDET in the condensed phase

A BABCψ ψ = ∆Assume Assume (A, B(A, B collective indices)collective indices)

due to the attractive interaction:due to the attractive interaction:

A B C† D†I ab ABCD a b aab b

*ABCD CDAB ABCD BACD ABDC

GL V4

V V , V V V

= − ε ε ψ ψ ψ ψ

= = =

DecomposeDecompose

24

( ) ( )AT B AB C† D* CD*int ABCD

CD* AT B

I cond

AB C† D*cond ABCD ABCD

int

G G

GL V

L V C

C C

V C4 4

L

4

L L

= Γ ψ ψ − Γ

= − ψ ψ − Γ ψ ψ

ψ

+ Γ

ψ

= +

CD * * CD*AB CDAB AB CDAB

G GV , V2 2

∆ = Γ ∆ = ΓWe defineWe define

* AT B A† B*cond AB AB

1 1L C C2 2

= ∆ ψ ψ − ∆ ψ ψ

and neglect and neglect LLint int . . ThereforeTherefore

25

A† B A† B * AT B A† BD AB AB

AB

dv 1L iV D iV D C C4 2 + + − − − + + − = ψ ⋅ ψ + ψ ⋅ ψ + ∆ ψ ψ − ∆ ψ ψ π ∑∫

(x) ( v,x)± +ψ ≡ ψ ±A

AA*

1C2

+

−

ψχ = ψ

NambuNambu--Gor’kovGor’kov basisbasis

AB ABA† BD *

AB AB

iV DdvLiV D4

⋅ −∆ = χ χ −∆ ⋅π

∫

† †

V1S( )(V )(V ) V

⋅ ∆= ⋅ ⋅ − ∆∆ ∆ ⋅

( )†, 0 ∆ ∆ =

26

* C† D*AB ABCD

G V C2

∆ = − ψ ψFrom the definition:From the definition:

one derives the gap equation (e.g. via functional one derives the gap equation (e.g. via functional formalism)formalism)

2 2* *AB ABCD CE2

ED

dv d 1iGV4 (2 ) D

µ∆ = ∆

π π π∫ ∫

†AB AB

1 1D (V )(V )

= ⋅ ⋅ − ∆∆

27

FourFour--fermifermi interaction oneinteraction one--gluon exchange inspiredgluon exchange inspired

a aI

3L G16

µµ= ψγ λ ψψγ λ ψ

8a a

a 1

ab dc ac bd

2( ) ( ) (3 )3

( ) ( ) 2

(1, ), (1, )

αβ δγ αγ βδ αβ γδ=

µµ

µ µ

λ λ = δ δ − δ δ

σ σ = ε ε

σ = σ σ = −σ

∑FierzFierz using:using:

† †I ( i)( j)( k)( ) i j k

( i)( j)( k)( ) ik j

GL V4

V (3 )

α β γ δα β γ δ

α β γ δ αδ βγ αγ βδ

= − ψ ψ ψ ψ

= − δ δ − δ δ δ δ

28

( i)( j) 3 ijα β αβ∆ = ε ε ∆In the 2SC caseIn the 2SC case

2 2

2 2 2 20

dv d4iG4 (2 )

µ ∆∆ =

π π π − − ∆∫ ∫

2

22 20

G d , 42

δ ∆ µ∆ = ρ ξ ρ =

πξ + ∆∫

4 pairing fermions4 pairing fermions

29

GG determined at determined at T = 0T = 0. . MM, constituent mass ~ , constituent mass ~ 400 400 MeVMeV

3

3 2 20

d p 11 8G(2 ) | p | M

Λ

=π +∫

Λ = µ + δwith with

ForFor 400 500 MeV, 800 MeV, M 400 MeVµ = − Λ = =

2SC 33 88 MeV∆ = −

Similar values for CFL.Similar values for CFL.

30

2

0 02 2

02 2 2

0 0

3 1 cosg 2 2(p ) dq d(cos )12 1 cos G /(2 )

1 1 cos (q )2 21 cos F /(2 ) q (q )

− θ∆ = θ +π − θ + µ

+ θ ∆

+ − θ + µ + ∆

∫ ∫

Gap equation in QCDGap equation in QCD

31

0q | q | 0→HardHard--loop approximationloop approximation

2 0D

qG m4 | q |π

=2 2

2D f 2

gm N2

µ=

π2DF m=

electricelectricmagneticmagnetic

For small For small momentamomentamagnetic gluons are magnetic gluons are unscreened and dominate unscreened and dominate giving a further giving a further logarithmic divergencelogarithmic divergence

20

0 02 2 20 0 0 0

g b (q )(p ) dq log18 | p q | q (q )

µ ∆∆ = π − + ∆

∫

5/ 24 5

f

2b 256 gN

− = π

32

2

0 00

32g

0

g b(p ) sin logp3 2

2b eπ

−

∆ ≈ ∆ π

∆ = µ

from the double logfrom the double log

To be trusted only for but, if To be trusted only for but, if extrapolated at 400extrapolated at 400--500 500 MeVMeV, gives values , gives values for the gap similar to the ones found using a for the gap similar to the ones found using a 44--fermi interaction.fermi interaction.

However condensation arises at asymptotic However condensation arises at asymptotic values of values of µ.µ.

510 GeVµ >

22 c / gg1 (log( / )) e

c−

≈ δ ∆ → ∆ ≈ δ

Results:Results:

33

Symmetries of Symmetries of superconducting phasessuperconducting phases

Consider 3 flavors, Consider 3 flavors, u,d,su,d,s.. AntisymmetryAntisymmetryrequiresrequires

kiL jL ijkAα β αβγ

γψ ψ = ε ε

DiagonalizeDiagonalize A through:A through:

c LSU(3) SU(3) DA g Ag A→ =

34

From the gap equations:From the gap equations:

(1,1,1) (1,1,0) (1,0,0)

1 0 0 1 0 0 1 0 0A 0 1 0 , A 0 1 0 , A 0 0 0

0 0 1 0 0 0 0 0 0

= = =

(1,1,1) (1,1,0) (1,0,0)∆ < ∆ < ∆with gapswith gaps

( 1,1,1) ( 1,1,0) ( 1,0,0)F F F< <butbut

since more fermions pair in the more since more fermions pair in the more symmetric configurations.symmetric configurations.

Confirmed by RG analysis (Evans et al.)Confirmed by RG analysis (Evans et al.)

35

From symmetry alone:From symmetry alone:* *

c L(R ) c L(R ) S c L(R ) c L(R )[(3 ,3 ) (3 ,3 )] (3 ,3 ) (6 ,6 )⊗ = ⊕

implying in generalimplying in generalI

iL jL ijI 6 i j i j

i j i j 6 i j i j

6 i j 6 i j

( )

( ) ( )

( ) ( )

α β αβ α β β α

α β β α α β β α

α β β α

ψ ψ = ∆ε ε + ∆ δ δ + δ δ =

= ∆ δ δ − δ δ + ∆ δ δ + δ δ =

= ∆ + ∆ δ δ + ∆ − ∆ δ δ

In NJL case with cutoff In NJL case with cutoff 800800 MeVMeV, , constituent mass constituent mass 400400 MeVMeV and and µ = µ = 400400 MeVMeV

685.3MeV, 1.3MeV∆ = ∆ = −

36

Original symmetry:Original symmetry:

QCD c L R B AG SU(3) SU(3) SU(3) U(1) U(1)= ⊗ ⊗ ⊗ ⊗

anomalousanomalousbroken tobroken toCFL c L R 2 2G SU(3) Z Z+ += ⊗ ⊗

anomalousanomalous

# Goldstones# Goldstones 3 8 1 8 8 81 11 17 1× + + − = + + + = +massivemassive

88 give mass to the gluons and give mass to the gluons and 8+18+1 are true are true masslessmassless Goldstone bosonsGoldstone bosons

37

Notice:Notice:

•• Breaking of Breaking of U(1)U(1)BB makes the makes the CFL phaseCFL phasesuperfluidsuperfluid

•• The CFL condensate is not gauge invariant, The CFL condensate is not gauge invariant, but considerbut consider

k ijk † k ijk †iL jL iR jRX ( ) , Y ( )α β α β

γ αβγ γ αβγ= ε ε ψ ψ = ε ε ψ ψ

i † i j * ij j(Y X) (Y ) Xα α

α

Σ = = ∑Σ Σ is gauge invariant and breaks the global part of is gauge invariant and breaks the global part of GGQCDQCD. . AlsoAlso

det(X), det(Y) 2 2B AU(1) U(1 Z Z) ⇒ ⊗⊗breakbreak

38

AU(1) is broken by the anomaly but induced by a 6is broken by the anomaly but induced by a 6--fermion operator irrelevant at the Fermi fermion operator irrelevant at the Fermi surface. Its contribution is parametrically surface. Its contribution is parametrically small and we expect a very light NGB small and we expect a very light NGB ((masslessmassless at infinite chemical potential)at infinite chemical potential)

Spectrum of the CFL phaseSpectrum of the CFL phase9

Ai A i

A 1

1 ( )2

α α

=

ψ = λ ψ∑Choose the basis:Choose the basis:

A

9 0

A 1, ,8 Gell Mann matrices

2 13

λ = ⋅⋅ ⋅ −

λ = λ = ×[ ]A B ABTr 2λ λ = δ

39

[ ]A iA i A

i

1 1( ) Tr2 2

αα

α

ψ = λ ψ = λ ψ∑Inverting:Inverting:

( )i j IA B ij

A B TA I B I

II

1 ( ) ( )2

Tr2

αβα β= λ λ ∆ε ε

∆ ψ ψ λ ε λ = ε

∑ ∑

I I( )αβ αβε = εT

I II

g g Tr[g]ε ε = −∑ for any 3x3 matrix gfor any 3x3 matrix g

A

A 1, ,8 AA 9 29

= ⋅⋅⋅ ∆ =∆∆ = = ∆ = − ∆

A BA ABψ ψ = ∆ δWe getWe get

2 2A A(p) (v )ε = ⋅ + ∆quasiquasi--fermionsfermions

40

gluonsgluons

2 2 2gm g F≈Expected Expected

NG coupling constantNG coupling constant

but wave function renormalization effects but wave function renormalization effects important (see later)important (see later)

NG bosonsNG bosons

Acquire mass through quark masses Acquire mass through quark masses except for the one related to the except for the one related to the breaking of breaking of U(1)U(1)BB

41

NG boson masses quadratic in NG boson masses quadratic in mmqq since the since the approximate invarianceapproximate invariance

2 L 2 R L(R ) L(R )(Z ) (Z ) :⊗ ψ → −ψ

L RM h.c.ψ ψ +and quark mass term:and quark mass term:

M M→ −

2 L 2 R(Z ) (Z )⊗Notice: anomaly breaks Notice: anomaly breaks through through instantonsinstantons, producing a , producing a chiralchiralcondensate (6 fermions condensate (6 fermions --> 2), but of > 2), but of order order ((ΛΛQCDQCD//µµ))88

42

InIn--medium electric chargemedium electric charge

aaD Qig T i Aµ µ µ µψ = ∂ ψ − ψ − ψ

The condensate breaks The condensate breaks U(1)U(1)em em but but leaves leaves invariant a combination ofinvariant a combination of Q Q andand TT88..

CFL vacuum:CFL vacuum: i i iX Yα α α= = δ

cSU(3) 82 1 1 2Q T diag , , Q

3 3 33 ≡ − = − − + =

Define:Define:

cSU(3)Q Q 1 1 Q Q 1 1 Q= ⊗ − ⊗ = ⊗ − ⊗ leaves invariant the leaves invariant the ground stateground state

i j jiQ X X Q Q Q 0αβ β α− → δ − δ =

43

Eigs(Q) 0, 1= ± Integers as in the HanIntegers as in the Han--NambuNambumodelmodel

8

A cos sin

g

A G

A sin cosGµ µ

µ

µ

µµ

= θ − θ

= θ + θ

rotated fieldsrotated fields

new interaction:new interaction:

( ) ( )

8s 8

ss

s

2

s

2

g g T 1 eA 1 Q eQA

2 e gtan , e ecos , gg cos3

3T cos Q 1 sin 1 Q2

g G Tµ µ µµ⊗ + ⊗ → +

θ = = θ =θ

= − θ ⊗ + θ ⊗

44

The The rotated “photon” remains rotated “photon” remains masslessmassless, , whereas the rotated gluons acquires a mass whereas the rotated gluons acquires a mass through the through the MeissnerMeissner effecteffect

A piece of CFL material for A piece of CFL material for masslessmassless quarks quarks would respond to an would respond to an emem field only through NGB:field only through NGB:

““bosonicbosonic metal”metal”

For quarks with equal masses, no light modes:For quarks with equal masses, no light modes:

“transparent insulator”“transparent insulator”

For different masses one needs non zero For different masses one needs non zero density of electrons or a density of electrons or a kaonkaon condensate condensate

leading to leading to masslessmassless excitationsexcitations

45

QuarkQuark--hadronhadron continuitycontinuity

CFL main properties:CFL main properties:

•• Integer charges (confinement)Integer charges (confinement)

•• Breaking of Breaking of chiralchiral symmetrysymmetry

•• Baryon number Baryon number superfluiditysuperfluidity

Except for last point as the Except for last point as the hadronichadronic phase. phase. However for 3However for 3--flavors a diflavors a di--baryon condensatebaryon condensate

H udsuds det[X]≈∼

is possible. For instance:is possible. For instance:

46

0 0 0(p ,n , , , )− + −Ξ Ξ Σ Σ Σ Σ ΛΛ

In this phase (In this phase (hypernuclearhypernuclear matter phase, HMPmatter phase, HMP) ) again the baryon number is broken and again the baryon number is broken and superfluiditysuperfluidity arises. This phase has the same arises. This phase has the same symmetries as CFL:symmetries as CFL:

No phase transition between No phase transition between CFL and HMP

(Schafer & (Schafer & WilczekWilczek))CFL and HMP

Why? Remember Why? Remember complementaritycomplementarity idea idea (Banks & (Banks & RabinoviciRabinovici; ; FradkinFradkin & & ShenkerShenker; ….); ….)

47

•• In some case two phases are In some case two phases are indistinguishable. From one phase to the other indistinguishable. From one phase to the other varying continuously the parameters. varying continuously the parameters.

•• Consider a confined and a broken phase:Consider a confined and a broken phase:

i R of G, states in the broken phase

Q R of G, states in t

elem

he confined phase

G G

entary

composi

,

te

R Rα

ψ ∈

∈

∼ ∼

Suppose there are Higgs fields such thatSuppose there are Higgs fields such that

iiQα α= ψ ϕ i i

α αϕ ≈ δ in the broken in the broken phase

withwithphase

48

Fermions in the two phases are the same but for Fermions in the two phases are the same but for a rea re--definition of the quantum numbersdefinition of the quantum numbers(neutrality of the Higgs in the broken phase)(neutrality of the Higgs in the broken phase)

Vector mesons in the broken phase from Vector mesons in the broken phase from gauge fields:gauge fields:

( )i * ij j(Z ) (g )β α

µ µ µ β α= −ϕ ∂ − ϕ

For CFL and HMP we have:For CFL and HMP we have: cG SU(3) , G SU(3)= =with fermions in the fundamental in three copieswith fermions in the fundamental in three copies

49

HYP phaseCFL phase

ikDα

αψ i ik kB Dα

α= ψ

j ki ijk i iD , Dα αβγ α α

β γ= ε ε ψ ψ ≈ δ

New definition of the electric charge in the New definition of the electric charge in the CFL phase. Add to the old generators the CFL phase. Add to the old generators the charges of the charges of the diquarkdiquark condensates. condensates. Quarks with the same charge as the Quarks with the same charge as the baryons. baryons. Vector mesonsVector mesons i i*

k kG D g Dα βα β=

50

Electric charges of the vector mesonsElectric charges of the vector mesons

001s

001d

-1-10u

sduikG

i i*k kG D g Dα β

α β=

51

•• In the HYP phase a In the HYP phase a nonetnonet of vector bosons, of vector bosons, but if the but if the dibaryondibaryon H exists the singlet vector H exists the singlet vector becomes unstable. Notice that in the CFL phase becomes unstable. Notice that in the CFL phase only 8 vector bosons.only 8 vector bosons.

•• In the CFL phase 8+1 quarks, but the gap of In the CFL phase 8+1 quarks, but the gap of the singlet is twice the one of the octet. The the singlet is twice the one of the octet. The singlet baryon has the structuresinglet baryon has the structure

and and H=SS

ijki j kS α β γ

αβγ= ε ε ψ ψ ψH=SS

52

Spectrum of the 2SC phaseSpectrum of the 2SC phase

( i)( j) 3 ijα β αβ∆ = ε ε ∆RememberRemember

No Goldstone bosonsNo Goldstone bosons

i3i

1,2 gapped

ungapped

αψ α =

ψ

c cSU(3) SU(2)

8 3 5 massive gluons

→

⇓− =

3i 3 gluons (M 0)ψ + =

(equal gap)(equal gap)

Light modes:Light modes:

53

2+1 flavors2+1 flavors

s u dm , m ,mµ ≈ µIt could happen It could happen

u d s

u d s

m ,m ,m

m ,m m

µ

⇓< µ <

Phase transition expectedPhase transition expected

2 2 2 2F F FE p M p M= µ = + → = µ −

The radius of The radius of the Fermi the Fermi sphere sphere decrasesdecraseswith the mass

1 2M M>with the mass

54

1 2

2 2F s Fp m , p= µ − = µSimple model:Simple model:

( ) ( )F F1 2p p3 3

2 2unpair s3 3

0 0

d p d p2 p m 2 | p |(2 ) (2 )

Ω = + − µ + −µπ π∫ ∫

( ) ( )F Fcomm commp p3 3 2 2

2 2pair s3 3 2

0 0

d p d p2 p m 2 | p |(2 ) (2 ) 4

µ ∆Ω = + − µ + −µ −

π π π∫ ∫condensation condensation

energyenergycomm

comm

2pair s

FF

m0 pp 4

∂Ω= → = µ −

∂ µ

( )4 2 2pair unpair s2

1 m 416

Ω − Ω ≈ − ∆ µπ

Condensation if:Condensation if:2sm

2µ >

∆

55

Notice that the Notice that the transition must be first transition must be first orderorder because for being in the pairing phasebecause for being in the pairing phase

2sm 0

2∆ > ≠

µ

(Minimal value of the gap to get condensation)(Minimal value of the gap to get condensation)

56

Effective Effective lagrangianslagrangians

•• Effective Effective lagrangianlagrangian for the CFL phase for the CFL phase

•• Effective Effective lagrangianlagrangian for the 2SC phasefor the 2SC phase

57

Effective Effective lagrangianlagrangian for the for the CFL phaseCFL phase

NG fields as the phases of the condensates in NG fields as the phases of the condensates in the representationthe representation(3, 3)

* *k ijk k ijkiL jL iR jRX , Yα β α β

γ αβγ γ αβγ= ε ε ψ ψ = ε ε ψ ψ

Quarks and X, Y transform asQuarks and X, Y transform asi( ) T i( ) T

L c L L R c R Ri i

c c L,R L,R B A

e g g , e g g

g SU(3) , g SU(3) , e U(1) , e U(1)

α+β α−β

α β

ψ → ψ ψ → ψ

∈ ∈ ∈ ∈

T 2i( ) T 2i( )c L c RX g Xg e , Y g Yg e− α+β − α−β→ →

58

)3(UY,X ∈SinceSinceThe number of NG fields isThe number of NG fields is

# X # Y (1 8) (1 8) 18+ = + + + =

8 of these fields give mass to the gluons. There 8 of these fields give mass to the gluons. There are onlyare only 10 physical NG bosons10 physical NG bosons corresponding to corresponding to

the breaking of the global symmetry (we consider the breaking of the global symmetry (we consider also the NGB associated to also the NGB associated to U(1)U(1)AA))

AVRL )1(U)1(U)3(SU)3(SU ⊗⊗⊗

22RL ZZ)3(SU ⊗⊗+

59

Better use fields belonging to SU(3). DefineBetter use fields belonging to SU(3). Define)(i2eXX θ+ϕ= )(i2eYY θ−ϕ=

andand)(i6

X e)Xdet(d θ+ϕ== )(i6Y e)Ydet(d θ−ϕ==TLc gXgX → T

Rc gYgY →transforming astransforming as

ϕ → ϕ − α θ → θ − β

The breaking of the global symmetry can be The breaking of the global symmetry can be described by the gauge invariant order parametersdescribed by the gauge invariant order parameters

i j ij X Y

ˆ ˆ ˆ ˆ(Y ) * X Y X, d , d+α αΣ = → Σ =

60

ΣΣ, , ddXX, , ddYY are 10 fields describing the physical NG are 10 fields describing the physical NG bosons. Alsobosons. Also

TLR gg Σ→Σ ∗

shows that shows that ΣΣTT transforms as thetransforms as the usual usual chiralchiral field.field.

Consider the currents:Consider the currents:† † † †

X

† † † †Y

as a

ˆ ˆ ˆ ˆ ˆ ˆ ˆJ XD X X( X X g ) X X gˆ ˆ ˆ ˆ ˆ ˆ ˆJ YD Y Y( Y Y g ) Y Y g

g ig g T

µ µ µ µ µ µ

µ µ µ µ µ µ

µ µ

= = ∂ + = ∂ +

= = ∂ + = ∂ +

=

†X,Y c X,Y cJ g J gµ µ→

61

Most general Most general lagrangianlagrangian up to two derivatives up to two derivatives invariant under invariant under G, the space rotation group O(3) G, the space rotation group O(3)

and Parityand Parity (R.C. & (R.C. & GattoGatto 1999)1999)

P : (X Y, , )↔ ϕ ↔ ϕ θ ↔ −θ

( ) ( ) ( ) ( )

( ) ( )

2 22 2 2 20 0 0 0T TX Y T X Y 0 0

22 2 22 22 20 0S SX Y S X Y

F F 1 1L Tr J J Tr J J4 4 2 2

vF F vTr J J Tr J J4 4 2 2

ϕ θ

= − − − α + + ∂ ϕ + ∂ θ +

+ − + α + − ∇ϕ − ∇θ

( ) ( )

( ) ( )

( ) ( )

2 22 2† † † †T T0 0 T 0 0 0

2 22 2† † † †S SS

2 22 22 20 0

F Fˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL Tr X X Y Y Tr X X Y Y 2g4 4

F Fˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆTr X X Y Y Tr X X Y Y 2g4 4

v1 1 v2 2 2 2

ϕ θ

= − ∂ − ∂ − α ∂ + ∂ + +

+ ∇ − ∇ + α ∇ + ∇ + +

+ ∂ ϕ + ∂ θ − ∇ϕ − ∇θ

62

Using SU(3)Using SU(3)cc gauge invariance we can choose:gauge invariance we can choose:a a

Ti T / F†ˆ ˆX Y e Π= =

Expanding at the second order in the fieldsExpanding at the second order in the fields

( ) ( ) ( )22 22 2 22 2 2

2 S

T

a a0 0 0

v1 1 1 v vL2 2 2

v

2

F

2F

2ϕ θ≈ ∂ Π + ∂ ϕ + ∂ θ − ∇Π − ∇ϕ − θ

=

∇

Gluons acquire Gluons acquire DebyeDebye and and MeissnerMeissner masses (not masses (not the rest masses, see later)the rest masses, see later)

2 2 2 2 2 2 2 2 2D T s T S S s S S s Tm g F , m g F v g F= α = α = α

63

Low energy theory supposed to be valid at Low energy theory supposed to be valid at energies << gap. Since we will see that gluons energies << gap. Since we will see that gluons

have masses of order have masses of order ∆∆ they can be decoupled they can be decoupled

( ) ( )

( ) ( )

( ) ( )

2 22 2† † † †T T0 0 T 0 0 0

2 22 2† † † †S ST

2 22 22 20 0

F Fˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆL Tr X X Y Y Tr X X Y Y 2g4 4

F Fˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆTr X X Y Y Tr X X Y Y 2g4 4

v1 1 v2 2 2 2

ϕ θ

= − ∂ − ∂ − α ∂ + ∂ + +

+ ∇ − ∇ + α ∇ + ∇ + +

+ ∂ ϕ + ∂ θ − ∇ϕ − ∇θ

( )† †1 ˆ ˆ ˆ ˆg X X Y Y2µ µ µ= − ∂ + ∂

64

we get the gauge invariant result:we get the gauge invariant result:

~ ~ χχ--lagrangianlagrangian

( )( ) ( )

22T

NGB t t

2 2 2 2 2 2t t

FL Tr v Tr4

1 1( ) v | | ( ) v | |2 2

+ +

ϕ θ

= ∂ Σ∂ Σ − ∇Σ∇Σ −

− ∂ ϕ − ∇ϕ − ∂ θ − ∇θ

ˆ ˆY X+Σ =

65

Effective Effective lagrangianlagrangian for the for the 2SC phase2SC phase

Light modesLight modes: 2 : 2 ungappedungapped fermions and 3 fermions and 3 masslessmasslessgluons of SU(2)gluons of SU(2)cc. Consider gluons:. Consider gluons:

Gauge invariance:Gauge invariance: a a a ai oi i ijk jk

1E F , B F2

= = ε

3a a a a

2a 1

1 1L E E B Bg 2 2=

ε = ⋅ − ⋅ λ ∑O(3)O(3) + Parity+ Parity

1v =ελ

Gluon velocityGluon velocity

66

Effective gauge coupling (Effective gauge coupling (λλ = 1= 1, see later):, see later):

2 2 22eff

Coul effg g gV gr r

= = ⇒ =ε ε

2 2 2' eff

s sg g g

4 c 4 v 4 cα = ⇒ α = =

π π π ε

Since Since ε >> 1 ε >> 1 (see later), gluons move slowly and(see later), gluons move slowly and

's

s

1 1α= <<

α ε

67

EquivelentlyEquivelently do the redo the re--scalingscaling0

0' ' '

'

0 0 1/ 4

a ' a '2

x gx , A A , g

L F1g

F

α α

µνµν

= = ε =εε

⇓

= −

2 2 2 2

2 2 2 2g g1

18 18µ µ

ε = + ≈π ∆ π ∆

Since from Since from 2

's

g4 c

α =π ε

's

3 g2 2

∆α =

µ

This defines the coupling at the matching scale This defines the coupling at the matching scale ∆∆

68

grows going to lower energies and becomes grows going to lower energies and becomes ~ 1 at The coupling is small at the ~ 1 at The coupling is small at the matching scale and we expect to be matching scale and we expect to be small. From onesmall. From one--loop betaloop beta--functionfunction

'sα

'QCDΛ

'QCDΛ

'0 s2 /'

QCD 02 2 22e exp ,

11 g 3− π β α π µ

Λ ≈ ∆ ≈ ∆ − β = ∆

25 3 / 2gc g e− − π∆ = µ

Numerical estimate very difficult, it Numerical estimate very difficult, it depends ondepends on

4c 512= π2

4 4c 512 exp8

+ π= π −

69

'QCD0.3 KeV 10 MeV≤ Λ ≤

QCD 200 MeV, 600 MeVΛ = µ =For increasing For increasing µ µ exponential decreasingexponential decreasing

'QCD

1Λ

Confinement radius grows Confinement radius grows exponentiallyexponentially

ForFor

Looking at fixed distance Looking at fixed distance and increasing the density and increasing the density there is a crossover at there is a crossover at which the color degrees of which the color degrees of freedom are unconfined.freedom are unconfined.